Lesson Area of a sector

Source code of 'Area of a sector'

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<H2>Area of a sector</H2>

<H3>Definition</H3>A &nbsp;<B>sector</B>&nbsp; of the circle is the plane figure restricted by the arc of the circle and by two radii connecting the center of the circle with endpoints of the arc &nbsp;(<B>Figure 1</B>).</H3>

A sector is characterized by two parameters: &nbsp;the radius of the circle and the central angle which leans on this arc.

<TABLE>
  <TR>
  <TD>
<H3>Theorem 1</H3>The area of a sector of the circle is half the product of the radius of the circle &nbsp;{{{r}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

and the radian measure of its central angle &nbsp;{{{alpha}}}

{{{S}}} = {{{1/2}}}{{{alpha}}}{{{r^2}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

<B>Proof</B>

Let us consider the ratio of the area of the sector &nbsp;{{{S}}}&nbsp; to the area &nbsp;{{{S[0]}}}&nbsp; of

the circle  of the same radius {{{r}}}, &nbsp;which is &nbsp;{{{S[0]}}} = {{{pi}}}{{{r^2}}}.

Based on the general properties of the area this ratio should be equal to &nbsp;{{{alpha/(2pi)}}}:

{{{S/S[0]}}} = {{{alpha/(2pi)}}}.

</TD>
  <TD>
{{{drawing( 180, 180,  -5.5, 5.5, -5.5, 5.5, 
            circle (0, 0, 0.1),
            locate(-0.2, 0, O),

circle (0, 0, 4),
        red(arc    (0, 0, 8.00, 8.00, 288, 360)),
        red(arc    (0, 0, 7.85, 7.85, 288, 360)),

arc    (0, 0, 8, 8,   0, 288),

arc    (0, 0, 2, 2, 288, 360),
            locate( 0.9, 1.2, alpha),

red(line (0, 0, 1.236, 3.804)),
        red(line (0, 0, 4.00, 0.00)),

locate( 1.8, 0.1, r),
            locate( 0.1, 2.6, r)
)}}}

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure 1</B>. A sector

</TD>
 </TR>
</TABLE>

From this proportion {{{S}}} = {{{alpha/(2pi)}}}.{{{S[0]}}} = {{{alpha/(2pi)}}}.{{{pi}}}{{{r^2}}} = {{{1/2}}}{{{alpha}}}{{{r^2}}}.

It is exactly what the formula &nbsp;(1)&nbsp; states.

<H3>Example 1</H3><TABLE>
  <TR>
  <TD>Find the area of a &nbsp;72�-sector of the circle which has the radius of &nbsp;10 cm.

<B>Solution</B>

Use the formula &nbsp;(1)&nbsp; above for the area of a sector. &nbsp;You have&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

{{{S}}} = {{{1/2}}}.{{{alpha}}}.{{{r^2}}} = {{{1/2}}}{{{(72*2*pi)/360}}}.{{{10^2}}} =

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= {{{1/2}}}*1.2566*100 = 62.832 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 180, 180,  -5.5, 5.5, -5.5, 5.5, 
            circle (0, 0, 0.1),
            locate(-0.2, 0, O),

circle (0, 0, 4),
        red(arc    (0, 0, 8.00, 8.00, 288, 360)),
        red(arc    (0, 0, 7.85, 7.85, 288, 360)),

arc    (0, 0, 8, 8,   0, 288),

arc    (0, 0, 2, 2, 288, 360),
            locate( 0.9, 1.2, alpha),

red(line (0, 0, 1.236, 3.804)),
        red(line (0, 0, 4.00, 0.00)),

locate( 1.8, 0.1, r),
            locate( 0.1, 2.6, r)
)}}}

<B>Figure 2</B>. To the <B>Example 1</B>

</TD>
 </TR>
</TABLE>

<B>Answer</B>. &nbsp;The area of the segment of the circle is &nbsp;62.832 {{{cm^2}}} (approximately).

<H3>Problem 1</H3><TABLE>
  <TR>
  <TD>Find the area of the figure which is restricted by the contour of a square&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

with the side of &nbsp;10 cm&nbsp; from the bottom and from the lateral sides and by

a semicircle from the top &nbsp;(<B>Figure 3a</B>).

<B>Solution</B>

Let us draw the horizontal line at the top of the square &nbsp;(<B>Figure 3b</B>).

Then our figure is the combination of the square with the side &nbsp;{{{a}}} = {{{10}}} {{{cm}}}

and the semicircle with the radius &nbsp;{{{r}}} = {{{5}}} {{{cm}}}.

The area of this combined figure is the sum of the area of the square

and the semicircle:

{{{S}}} = {{{a^2}}} + {{{1/2}}}.{{{pi}}}{{{r^2}}} = {{{10^2}}} + {{{1/2}}}*{{{3.14159}}}*{{{5^2}}} = 100 + 39.27 =

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= 139.27 {{{cm^2}}} (approximately).

<B>Answer</B>. &nbsp;The area of the combined figure is &nbsp;139.27 {{{cm^2}}} (approximately).

</TD>
  <TD>
{{{drawing( 160, 200,  -4.0, 4.0, -3.7, 6.3,

line (-3, -3, 3, -3),
            line (-3, -3,-3,  3),
            line ( 3, -3, 3,  3),

arc    (0, 3, 6.00, 6.00, 180, 360),

circle (0, 3, 0.1),
            locate(-0.2, 2.9, O),

locate(-3.2, -3,   A),
            locate( 2.9, -3,   B),
            locate( 3.1,  3.3, C),
            locate(-3.5,  3.3, D)
)}}}

<B>Figure 3a</B>. To the <B>Problem 1</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD>
{{{drawing( 160, 200,  -4.0, 4.0, -3.7, 6.3,

line (-3, -3, 3, -3),
            line (-3, -3,-3,  3),
            line ( 3, -3, 3,  3),

arc    (0, 3, 6.00, 6.00, 180, 360),

circle (0, 3, 0.1),
            locate(-0.2, 2.9, O),

locate(-3.2, -3,   A),
            locate( 2.9, -3,   B),
            locate( 3.1,  3.3, C),
            locate(-3.5,  3.3, D),

green(line (-3,   3, 3,  3)),
            locate(1.3, 3.9, r)

)}}}

<B>Figure 3b</B>. &nbsp;To the solution
&nbsp;&nbsp;&nbsp;&nbsp;of the <B>Problem 1</B>

</TD>
 </TR>
</TABLE>

<H3>Problem 2</H3><TABLE>
  <TR>
  <TD>Find the area of the figure which is obtained from a square with the side&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

of &nbsp;10 cm&nbsp; after cutting off a semicircle of the radius of &nbsp;6 cm&nbsp; as shown

in the &nbsp;<B>Figure 4</B>.

<B>Solution</B>

The area of the figure under consideration is the difference of the area

of the square with the side &nbsp;{{{a}}} = {{{10}}} {{{cm}}}&nbsp; and the semicircle with the

radius &nbsp;{{{r}}} = {{{3}}} {{{cm}}}. &nbsp;So,

{{{S}}} = {{{a^2}}} - {{{1/2}}}.{{{pi}}}{{{r^2}}} = {{{10^2}}} - {{{1/2}}}*{{{3.14159}}}*{{{3^2}}} = 100 - 14.14 =

</TD>
  <TD>
{{{drawing( 160, 200,  -4.0, 4.0, -3.7, 6.3,

line (-3, -3, 3, -3),
            line (-3, -3,-3,  3),
            line ( 3, -3, 3,  3),
            line (-3,  3, 3,  3),

line (-2,  1, 2,  1),
            arc    (0, 1, 4.00, 4.00, 0, 180),

circle (0,   1,   0.1),
            locate( 0.2, 0.9, O),

locate(-3.2, -3,   A),
            locate( 2.9, -3,   B),
            locate( 3.1,  3.3, C),
            locate(-3.5,  3.3, D)
)}}}

<B>Figure 4</B>. To the <B>Problem 2</B>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
 </TR>
</TABLE>

<B>Answer</B>. &nbsp;The area of the figure under consideration is &nbsp;85.86 {{{cm^2}}} (approximately).

My other lessons on the area of a circle, &nbsp;the area of a sector and the area of a segment of the circle in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-circle.lesson>Area of a circle</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Surface-area/Area-of-a-segment-of-the-circle.lesson>Area of a segment of the circle</A>

under the current topic &nbsp;<B>Area and surface area</B>&nbsp; of the section &nbsp;<B>Geometry</B>, &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-a-circle.lesson>Solved problems on area of a circle</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-a-sector.lesson>Solved problems on area of a sector</A>,

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-a-segment-of-the-circle.lesson>Solved problems on area of a segment of the circle</A> &nbsp;and

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-area-of-aCircle-aSector-and-aSegment-of-the-circle.lesson>Solved problems on area of a circle, a sector and a segment of the circle</A>

under the topic &nbsp;<B>Geometry</B>&nbsp; of the section &nbsp;<B>Word problems</B>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.